Solution 2.3:3d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 2.3:3d moved to Solution 2.3:3d: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
Because both terms,
-
<center> [[Image:2_3_3d.gif]] </center>
+
<math>x\left( x+3 \right)</math>
-
{{NAVCONTENT_STOP}}
+
and
 +
<math>x\left( 2x-9 \right)</math>
 +
contain the factor
 +
<math>x</math>, we can take out
 +
<math>x</math> from the left-hand side and collect together the remaining expression:
 +
 
 +
 
 +
<math>\begin{align}
 +
& x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\
 +
& =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\
 +
\end{align}</math>
 +
 
 +
 
 +
The equation is thus
 +
 
 +
 
 +
<math>x\left( -x+12 \right)=0</math>
 +
 
 +
and we obtain directly that the equation is satisfied if either
 +
<math>x</math>
 +
or
 +
<math>-x+\text{12}</math>
 +
is zero. The solutions to the equation are therefore
 +
<math>x=0\text{ }</math>
 +
and
 +
<math>x=\text{12}</math>.
 +
 
 +
Here, it can be worth checking that
 +
<math>x=\text{12 }</math>
 +
is a solution (the case
 +
<math>x=0</math>
 +
is obvious):
 +
 
 +
LHS
 +
<math>=12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0=</math>
 +
RHS

Revision as of 15:05, 20 September 2008

Because both terms, \displaystyle x\left( x+3 \right) and \displaystyle x\left( 2x-9 \right) contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression:


\displaystyle \begin{align} & x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\ & =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\ \end{align}


The equation is thus


\displaystyle x\left( -x+12 \right)=0

and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+\text{12} is zero. The solutions to the equation are therefore \displaystyle x=0\text{ } and \displaystyle x=\text{12}.

Here, it can be worth checking that \displaystyle x=\text{12 } is a solution (the case \displaystyle x=0 is obvious):

LHS \displaystyle =12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0= RHS

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_2.3:3d"